Download Document

Physics 7C lecture 03
Application of Newton’s
laws: free body diagram
Thursday October 3, 8:00 AM – 9:20 AM
Engineering Hall 1200
Copyright © 2012 Pearson Education Inc.
Newton’s laws
• Newton’s First Law of Motion: when forces are balanced
the object will keep its motion/velocity.
• Newton’s Second Law of Motion: F = m a, force causes
changes to the motion/velocity.
• Newton’s Third Law of Motion: action-reaction pairs
• Yet: The laws are simple to state but intricate in their
application.
• Let’s do some iClicker questions first.
Copyright © 2012 Pearson Education Inc.
Q4.1
An elevator is being lifted at a constant
speed by a steel cable attached to an electric
motor. There is no air resistance, nor is
there any friction between the elevator and Cable
the walls of the elevator shaft.
The upward force exerted on the elevator
by the cable is
Motor
v
Elevator
A. greater than the downward force of gravity.
B. equal to the force of gravity.
C. less than the force of gravity.
D. any of the above, depending on the speed of the elevator.
© 2012 Pearson Education, Inc.
A4.1
An elevator is being lifted at a constant
speed by a steel cable attached to an electric
motor. There is no air resistance, nor is
there any friction between the elevator and Cable
the walls of the elevator shaft.
The upward force exerted on the elevator
by the cable is
Motor
v
Elevator
A. greater than the downward force of gravity.
B. equal to the force of gravity.
C. less than the force of gravity.
D. any of the above, depending on the speed of the elevator.
© 2012 Pearson Education, Inc.
Q4.2
An elevator is being lowered at a constant
speed by a steel cable attached to an electric
motor. There is no air resistance, nor is
there any friction between the elevator and Cable
the walls of the elevator shaft.
The upward force exerted on the elevator
by the cable is
Motor
v
Elevator
A. greater than the downward force of gravity.
B. equal to the force of gravity.
C. less than the force of gravity.
D. any of the above, depending on the speed of the elevator.
© 2012 Pearson Education, Inc.
A4.2
An elevator is being lowered at a constant
speed by a steel cable attached to an electric
motor. There is no air resistance, nor is
there any friction between the elevator and Cable
the walls of the elevator shaft.
The upward force exerted on the elevator
by the cable is
Motor
v
Elevator
A. greater than the downward force of gravity.
B. equal to the force of gravity.
C. less than the force of gravity.
D. any of the above, depending on the speed of the elevator.
© 2012 Pearson Education, Inc.
Q4.5
A lightweight crate (A) and a heavy
crate (B) are side-by-side on a
F
horizontal floor. You apply a
horizontal force F to crate A. There is
friction between the crates and the
floor.
If the two crates are accelerating to the right,
A
B
A. crate A exerts more force on crate B than B exerts on A
B. crate A exerts less force on crate B than B exerts on A
C. crate A exerts as much force on crate B as B exerts on A
D. Answer depends on the details of the friction force
© 2012 Pearson Education, Inc.
A4.5
A lightweight crate (A) and a heavy
crate (B) are side-by-side on a
F
horizontal floor. You apply a
horizontal force F to crate A. There is
friction between the crates and the
floor.
If the two crates are accelerating to the right,
A
B
A. crate A exerts more force on crate B than B exerts on A
B. crate A exerts less force on crate B than B exerts on A
C. crate A exerts as much force on crate B as B exerts on A
D. Answer depends on the details of the friction force
© 2012 Pearson Education, Inc.
Q4.6
An elevator is being lowered at constant
speed by a steel cable attached to an electric
motor. There is no air resistance, nor is
Cable
there any friction between the elevator and
the walls of the elevator shaft.
The upward force exerted on the elevator
by the cable has the same magnitude as the
force of gravity on the elevator, but points
in the opposite direction. Why?
A. Newton’s first law
B. Newton’s second law
C. Newton’s third law
© 2012 Pearson Education, Inc.
Motor
v
Elevator
A4.6
An elevator is being lowered at constant
speed by a steel cable attached to an electric
motor. There is no air resistance, nor is
Cable
there any friction between the elevator and
the walls of the elevator shaft.
The upward force exerted on the elevator
by the cable has the same magnitude as the
force of gravity on the elevator, but points
in the opposite direction. Why?
A. Newton’s first law
B. Newton’s second law
C. Newton’s third law
© 2012 Pearson Education, Inc.
Motor
v
Elevator
Q4.3
vx
The graph to the right shows the
velocity of an object as a function
of time.
Which of the graphs below best
shows the net force versus time
for this object?
Fx
Fx
t
0
A.
© 2012 Pearson Education, Inc.
Fx
t
0
B.
t
0
Fx
t
0
C.
Fx
t
0
D.
t
0
E.
A4.3
vx
The graph to the right shows the
velocity of an object as a function
of time.
Which of the graphs below best
shows the net force versus time
for this object?
Fx
Fx
t
0
A.
© 2012 Pearson Education, Inc.
Fx
t
0
B.
t
0
Fx
t
0
C.
Fx
t
0
D.
t
0
E.
Free-body diagrams
• A free-body diagram is a sketch showing all the forces acting
on an object.
• Why do we need it?
What is the tension force in the rope?
Copyright © 2012 Pearson Education Inc.
Free-body diagrams
• We need to isolate the objects and study the forces on each of
them.
T1
T2
W1
T1
W2
T2
W3
Note: W = m g
action-reaction pairs
Copyright © 2012 Pearson Education Inc.
Free-body diagrams
• We can apply the 2nd law to each of the following, or to the
group.
T1
T2
W1
T1
W2
W3
T2
The three objects share the same acceleration:
a = (F-W1-W2-W3) / (m1+m2+m3)
= F/(m1+m2+m3) – g = 200/(6+4+5) – 9.8 = 3.53 m/s2
Copyright © 2012 Pearson Education Inc.
Free-body diagrams
• Now we focus on m1 and solve for tension T1.
for the upper mass, apply
the 2nd law:
F-W1-T1 = m1 a
we have:
W1
T1
Copyright © 2012 Pearson Education Inc.
T1
= F-W1-m1 a
= 200-6*9.8-6*3.53
= 120.02 N
Free-body diagrams
• Similarly we focus on the rope and solve for tension T2.
T1-W2-T2 = m2 a
we have:
T1
T2
= T1-W2-m2 a
= 120.02-4*9.8-4*3.53 W2
= 66.7 N
T2
Copyright © 2012 Pearson Education Inc.
Free-body diagrams
• We can double check the result on the lower mass:
T2
T2-W3 = m3 a
we have calculated:
T2 = 66.7 N
a = 3.53 m/s2
66.7 – 5*9.8 = 5 * 3.53
true!
Copyright © 2012 Pearson Education Inc.
W3
?
Free-body diagrams—examples
• A free-body diagram is a sketch showing all the forces acting
on an object.
Copyright © 2012 Pearson Education Inc.
Free-body diagrams—examples
• A free-body diagram is a sketch showing all the forces acting
on an object.
Copyright © 2012 Pearson Education Inc.
Free-body diagrams—examples
Copyright © 2012 Pearson Education Inc.
Q4.9
A person pulls horizontally on
block B, causing both blocks to
move horizontally as a unit.
There is friction between block B
and the horizontal table.
If the two blocks are moving to
the right at constant velocity,
A. the horizontal force that B exerts on A points to the left.
B. the horizontal force that B exerts on A points to the right.
C. B exerts no horizontal force on A.
D. not enough information given to decide
© 2012 Pearson Education, Inc.
A4.9
A person pulls horizontally on
block B, causing both blocks to
move horizontally as a unit.
There is friction between block B
and the horizontal table.
If the two blocks are moving to
the right at constant velocity,
A. the horizontal force that B exerts on A points to the left.
B. the horizontal force that B exerts on A points to the right.
C. B exerts no horizontal force on A.
D. not enough information given to decide
© 2012 Pearson Education, Inc.
Q4.12
A woman pulls on a 6.00kg crate, which in turn is
connected to a 4.00-kg
crate by a light rope. The
light rope remains taut.
Compared to the 6.00-kg crate, the lighter 4.00-kg crate
A. is subjected to the same net force and has the same acceleration.
B. is subjected to a smaller net force and has the same acceleration.
C. is subjected to the same net force and has a smaller acceleration.
D. is subjected to a smaller net force and has a smaller acceleration.
E. none of the above
© 2012 Pearson Education, Inc.
A4.12
A woman pulls on a 6.00kg crate, which in turn is
connected to a 4.00-kg
crate by a light rope. The
light rope remains taut.
Compared to the 6.00-kg crate, the lighter 4.00-kg crate
A. is subjected to the same net force and has the same acceleration.
B. is subjected to a smaller net force and has the same acceleration.
C. is subjected to the same net force and has a smaller acceleration.
D. is subjected to a smaller net force and has a smaller acceleration.
E. none of the above
© 2012 Pearson Education, Inc.
Q5.1
A car engine is suspended from a chain linked at O to two
other chains. Which of the following forces should be
included in the free-body diagram for the engine?
A. tension T1
B. tension T2
C. tension T3
D. two of the above
E. T1, T2, and T3
© 2012 Pearson Education, Inc.
A5.1
A car engine is suspended from a chain linked at O to two
other chains. Which of the following forces should be
included in the free-body diagram for the engine?
A. tension T1
B. tension T2
C. tension T3
D. two of the above
E. T1, T2, and T3
© 2012 Pearson Education, Inc.
Two-dimensional equilibrium
• A car engine hangs from several chains.
Copyright © 2012 Pearson Education Inc.
Two-dimensional equilibrium
• A car engine hangs from several chains.
Copyright © 2012 Pearson Education Inc.
Q5.3
A cart (weight w1) is attached
by a lightweight cable to a
bucket (weight w2) as shown.
The ramp is frictionless.
When released, the cart accelerates up the ramp.
Which of the following is a correct free-body diagram for the cart?
n
n
n
n
T
T
T
T
w1
A.
© 2012 Pearson Education, Inc.
w1
B.
w1
m1a
C.
w1
m1a
D.
A5.3
A cart (weight w1) is attached
by a lightweight cable to a
bucket (weight w2) as shown.
The ramp is frictionless.
When released, the cart accelerates up the ramp.
Which of the following is a correct free-body diagram for the cart?
n
n
n
n
T
T
T
T
w1
A.
© 2012 Pearson Education, Inc.
w1
B.
w1
m1a
C.
w1
m1a
D.
Q5.4
A cart (weight w1) is attached by
a lightweight cable to a bucket
(weight w2) as shown. The ramp
is frictionless. The pulley is
frictionless and does not rotate.
When released, the cart accelerates up the ramp and the bucket
accelerates downward. How does the cable tension T compare to w2?
A. T = w2
B. T > w2
C. T < w2
D. not enough information given to decide
© 2012 Pearson Education, Inc.
A5.4
A cart (weight w1) is attached by
a lightweight cable to a bucket
(weight w2) as shown. The ramp
is frictionless. The pulley is
frictionless and does not rotate.
When released, the cart accelerates up the ramp and the bucket
accelerates downward. How does the cable tension T compare to w2?
A. T = w2
B. T > w2
C. T < w2
D. not enough information given to decide
© 2012 Pearson Education, Inc.
Bodies connected by a cable and pulley
• If the cart is moving at constant speed, what is w2/w1?
Copyright © 2012 Pearson Education Inc.
Bodies connected by a cable and pulley
• If the cart is moving at constant speed, what is w2/w1?
consider the bucket
T = w2
Copyright © 2012 Pearson Education Inc.
Bodies connected by a cable and pulley
• If the cart is moving at constant speed, what is w2/w1?
consider the cart along x direction:
W1 Sin 15o – T = 0
Thus: W2 = W1 Sin 15o
and: W2/W1 = Sin 15o = 0.26
Copyright © 2012 Pearson Education Inc.